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Capacitance

165998 The capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value. If a dielectric slab of thickness $t = \frac{d}{2}$ is inserted between the plates (where, $d$ is the distance of separation between the plates). What is the dielectric constant of the slab?

1

K = 2

2

K = \frac{1}{2}

3

K = 1

4

K = \sqrt{2}

Explanation:

: Given, New capacitance is $\frac{4}{3}$ of original,
$C_{2} = \frac{4}{3} C_{1}$
$1^{st} condition \to capacitor in Air (without Dielectric)$
$C_{1} = \frac{ε_{0} A}{d}$
$\underset{d}{air}$
$C_{2} = \frac{ε_{0} A}{\frac{d}{2 K} + \frac{d}{2}} = \frac{2 ε_{0} A}{d (\frac{1}{K} + 1)}$
$\frac{2 ε_{0} A}{d (\frac{1}{K} + 1)} = \frac{4}{3} \times \frac{ε_{0} A}{d} [From equation (i)]$
$\frac{2}{\frac{1}{K} + 1} = \frac{4}{3}$
$\frac{1}{K} + 1 = \frac{3}{2}$
$\frac{1}{K} = \frac{1}{2}$
$K = 2$

[JCECE-2017]

Capacitance

165999 A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5 , the percentage increase in the capacitance will be

1

400 %

2

66.6 %

3

33.3 %

4

200 %

Explanation:

: Given, Initial Capacitance,
$C = \frac{ε_{0} A}{d}$
Now, Half filled with dielectric of constant, $K = 5$ Then New capacitance,
$C_{2} = \frac{ε_{0} A}{(\frac{d}{2} + \frac{d / 2}{K})}$
$C_{2} = \frac{ε_{0} A}{d} \times \frac{2}{1 + \frac{1}{K}}$
$C_{2} = \frac{ε_{0} A}{d} \times \frac{2}{1 + \frac{1}{5}}$
$C_{2} = \frac{5}{3} C$
$∴ %$ increased in capacitance, $= \frac{5}{3} - 1 = \frac{2}{3} \times 100$
$= 66.6 %$
So, increase in the capacitance is $66.6 %$

[JCECE-2009]

Capacitance

166000 The capacitors $A$ and $B$ have identical geometry. A material with a dielectric constant 3 is present between the plates of $B$ . The potential difference across $A$ and $B$ are respectively:

1

2.5 V, 7.5 V

2

2 V, 8 V

3

8 V, 2 V

4

7.5 V, 2.5 V

Explanation:

: Let's potential difference across A and B are $V_{A}$ and $V_{B}$ .
We know that, $Q = CV$
Then, $C \propto \frac{1}{V}$
And capacitance,
$C_{A} = C$
$C_{B} = 3 C$
$\frac{V_{A}}{V_{B}} = \frac{C_{B}}{C_{A}}$
$\frac{V_{A}}{V_{B}} = \frac{3 C}{C}$
$V_{A} = 3 V_{B}$
$(∵ K = 3)$
We know that,
$\begin{array}{ll} V_{A} + V_{B} = 10 \\ 3 V_{B} + V_{B} = 10 \\ 4 V_{B} = 10 \Rightarrow V_{B} = 2.5 V \\ ∴ V_{A} = 3 \times 2.5 = 7.5 V \end{array}$

[JCECE-2005]

Capacitance

166001 A parallel plate capacitor has a capacity $C$ . The separation between the plates is doubled and a dielectric medium is inserted between the plates. If the capacity is $3 C$ , then the dielectric constant of the medium will be

1 1.5

2 3

3 6

4 12

Explanation:

: Given, $C_{1} = C, C_{2} = 3 C$
Let's $d_{1} = d,$ so, $d_{2} = 2 d$
Then, dielectric constant $(K) =$ ?
$C_{1} = \frac{ε_{0} A}{d_{1}} = \frac{ε_{0} A}{d}$
Now, $C_{2} = \frac{K ε_{0} A}{d_{2}}$
$3 C = \frac{K ε_{0} A}{2 d}$
$3 C = \frac{K}{2} \times \frac{ε_{0} A}{d}$
$3 C = \frac{K}{2} C_{1} \Rightarrow 3 C = \frac{K}{2} C$
$K = 6$

[COMEDK 2015]

Capacitance

165998 The capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value. If a dielectric slab of thickness $t = \frac{d}{2}$ is inserted between the plates (where, $d$ is the distance of separation between the plates). What is the dielectric constant of the slab?

1

K = 2

2

K = \frac{1}{2}

3

K = 1

4

K = \sqrt{2}

Explanation:

: Given, New capacitance is $\frac{4}{3}$ of original,
$C_{2} = \frac{4}{3} C_{1}$
$1^{st} condition \to capacitor in Air (without Dielectric)$
$C_{1} = \frac{ε_{0} A}{d}$
$\underset{d}{air}$
$C_{2} = \frac{ε_{0} A}{\frac{d}{2 K} + \frac{d}{2}} = \frac{2 ε_{0} A}{d (\frac{1}{K} + 1)}$
$\frac{2 ε_{0} A}{d (\frac{1}{K} + 1)} = \frac{4}{3} \times \frac{ε_{0} A}{d} [From equation (i)]$
$\frac{2}{\frac{1}{K} + 1} = \frac{4}{3}$
$\frac{1}{K} + 1 = \frac{3}{2}$
$\frac{1}{K} = \frac{1}{2}$
$K = 2$

[JCECE-2017]

Capacitance

165999 A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5 , the percentage increase in the capacitance will be

1

400 %

2

66.6 %

3

33.3 %

4

200 %

Explanation:

: Given, Initial Capacitance,
$C = \frac{ε_{0} A}{d}$
Now, Half filled with dielectric of constant, $K = 5$ Then New capacitance,
$C_{2} = \frac{ε_{0} A}{(\frac{d}{2} + \frac{d / 2}{K})}$
$C_{2} = \frac{ε_{0} A}{d} \times \frac{2}{1 + \frac{1}{K}}$
$C_{2} = \frac{ε_{0} A}{d} \times \frac{2}{1 + \frac{1}{5}}$
$C_{2} = \frac{5}{3} C$
$∴ %$ increased in capacitance, $= \frac{5}{3} - 1 = \frac{2}{3} \times 100$
$= 66.6 %$
So, increase in the capacitance is $66.6 %$

[JCECE-2009]

Capacitance

166000 The capacitors $A$ and $B$ have identical geometry. A material with a dielectric constant 3 is present between the plates of $B$ . The potential difference across $A$ and $B$ are respectively:

1

2.5 V, 7.5 V

2

2 V, 8 V

3

8 V, 2 V

4

7.5 V, 2.5 V

Explanation:

: Let's potential difference across A and B are $V_{A}$ and $V_{B}$ .
We know that, $Q = CV$
Then, $C \propto \frac{1}{V}$
And capacitance,
$C_{A} = C$
$C_{B} = 3 C$
$\frac{V_{A}}{V_{B}} = \frac{C_{B}}{C_{A}}$
$\frac{V_{A}}{V_{B}} = \frac{3 C}{C}$
$V_{A} = 3 V_{B}$
$(∵ K = 3)$
We know that,
$\begin{array}{ll} V_{A} + V_{B} = 10 \\ 3 V_{B} + V_{B} = 10 \\ 4 V_{B} = 10 \Rightarrow V_{B} = 2.5 V \\ ∴ V_{A} = 3 \times 2.5 = 7.5 V \end{array}$

[JCECE-2005]

Capacitance

166001 A parallel plate capacitor has a capacity $C$ . The separation between the plates is doubled and a dielectric medium is inserted between the plates. If the capacity is $3 C$ , then the dielectric constant of the medium will be

1 1.5

2 3

3 6

4 12

Explanation:

: Given, $C_{1} = C, C_{2} = 3 C$
Let's $d_{1} = d,$ so, $d_{2} = 2 d$
Then, dielectric constant $(K) =$ ?
$C_{1} = \frac{ε_{0} A}{d_{1}} = \frac{ε_{0} A}{d}$
Now, $C_{2} = \frac{K ε_{0} A}{d_{2}}$
$3 C = \frac{K ε_{0} A}{2 d}$
$3 C = \frac{K}{2} \times \frac{ε_{0} A}{d}$
$3 C = \frac{K}{2} C_{1} \Rightarrow 3 C = \frac{K}{2} C$
$K = 6$

[COMEDK 2015]

Capacitance

165998 The capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value. If a dielectric slab of thickness $t = \frac{d}{2}$ is inserted between the plates (where, $d$ is the distance of separation between the plates). What is the dielectric constant of the slab?

1

K = 2

2

K = \frac{1}{2}

3

K = 1

4

K = \sqrt{2}

Explanation:

: Given, New capacitance is $\frac{4}{3}$ of original,
$C_{2} = \frac{4}{3} C_{1}$
$1^{st} condition \to capacitor in Air (without Dielectric)$
$C_{1} = \frac{ε_{0} A}{d}$
$\underset{d}{air}$
$C_{2} = \frac{ε_{0} A}{\frac{d}{2 K} + \frac{d}{2}} = \frac{2 ε_{0} A}{d (\frac{1}{K} + 1)}$
$\frac{2 ε_{0} A}{d (\frac{1}{K} + 1)} = \frac{4}{3} \times \frac{ε_{0} A}{d} [From equation (i)]$
$\frac{2}{\frac{1}{K} + 1} = \frac{4}{3}$
$\frac{1}{K} + 1 = \frac{3}{2}$
$\frac{1}{K} = \frac{1}{2}$
$K = 2$

[JCECE-2017]

Capacitance

165999 A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5 , the percentage increase in the capacitance will be

1

400 %

2

66.6 %

3

33.3 %

4

200 %

Explanation:

: Given, Initial Capacitance,
$C = \frac{ε_{0} A}{d}$
Now, Half filled with dielectric of constant, $K = 5$ Then New capacitance,
$C_{2} = \frac{ε_{0} A}{(\frac{d}{2} + \frac{d / 2}{K})}$
$C_{2} = \frac{ε_{0} A}{d} \times \frac{2}{1 + \frac{1}{K}}$
$C_{2} = \frac{ε_{0} A}{d} \times \frac{2}{1 + \frac{1}{5}}$
$C_{2} = \frac{5}{3} C$
$∴ %$ increased in capacitance, $= \frac{5}{3} - 1 = \frac{2}{3} \times 100$
$= 66.6 %$
So, increase in the capacitance is $66.6 %$

[JCECE-2009]

Capacitance

166000 The capacitors $A$ and $B$ have identical geometry. A material with a dielectric constant 3 is present between the plates of $B$ . The potential difference across $A$ and $B$ are respectively:

1

2.5 V, 7.5 V

2

2 V, 8 V

3

8 V, 2 V

4

7.5 V, 2.5 V

Explanation:

: Let's potential difference across A and B are $V_{A}$ and $V_{B}$ .
We know that, $Q = CV$
Then, $C \propto \frac{1}{V}$
And capacitance,
$C_{A} = C$
$C_{B} = 3 C$
$\frac{V_{A}}{V_{B}} = \frac{C_{B}}{C_{A}}$
$\frac{V_{A}}{V_{B}} = \frac{3 C}{C}$
$V_{A} = 3 V_{B}$
$(∵ K = 3)$
We know that,
$\begin{array}{ll} V_{A} + V_{B} = 10 \\ 3 V_{B} + V_{B} = 10 \\ 4 V_{B} = 10 \Rightarrow V_{B} = 2.5 V \\ ∴ V_{A} = 3 \times 2.5 = 7.5 V \end{array}$

[JCECE-2005]

Capacitance

166001 A parallel plate capacitor has a capacity $C$ . The separation between the plates is doubled and a dielectric medium is inserted between the plates. If the capacity is $3 C$ , then the dielectric constant of the medium will be

1 1.5

2 3

3 6

4 12

Explanation:

: Given, $C_{1} = C, C_{2} = 3 C$
Let's $d_{1} = d,$ so, $d_{2} = 2 d$
Then, dielectric constant $(K) =$ ?
$C_{1} = \frac{ε_{0} A}{d_{1}} = \frac{ε_{0} A}{d}$
Now, $C_{2} = \frac{K ε_{0} A}{d_{2}}$
$3 C = \frac{K ε_{0} A}{2 d}$
$3 C = \frac{K}{2} \times \frac{ε_{0} A}{d}$
$3 C = \frac{K}{2} C_{1} \Rightarrow 3 C = \frac{K}{2} C$
$K = 6$

[COMEDK 2015]

Capacitance

165998 The capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value. If a dielectric slab of thickness $t = \frac{d}{2}$ is inserted between the plates (where, $d$ is the distance of separation between the plates). What is the dielectric constant of the slab?

1

K = 2

2

K = \frac{1}{2}

3

K = 1

4

K = \sqrt{2}

Explanation:

: Given, New capacitance is $\frac{4}{3}$ of original,
$C_{2} = \frac{4}{3} C_{1}$
$1^{st} condition \to capacitor in Air (without Dielectric)$
$C_{1} = \frac{ε_{0} A}{d}$
$\underset{d}{air}$
$C_{2} = \frac{ε_{0} A}{\frac{d}{2 K} + \frac{d}{2}} = \frac{2 ε_{0} A}{d (\frac{1}{K} + 1)}$
$\frac{2 ε_{0} A}{d (\frac{1}{K} + 1)} = \frac{4}{3} \times \frac{ε_{0} A}{d} [From equation (i)]$
$\frac{2}{\frac{1}{K} + 1} = \frac{4}{3}$
$\frac{1}{K} + 1 = \frac{3}{2}$
$\frac{1}{K} = \frac{1}{2}$
$K = 2$

[JCECE-2017]

Capacitance

165999 A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5 , the percentage increase in the capacitance will be

1

400 %

2

66.6 %

3

33.3 %

4

200 %

Explanation:

: Given, Initial Capacitance,
$C = \frac{ε_{0} A}{d}$
Now, Half filled with dielectric of constant, $K = 5$ Then New capacitance,
$C_{2} = \frac{ε_{0} A}{(\frac{d}{2} + \frac{d / 2}{K})}$
$C_{2} = \frac{ε_{0} A}{d} \times \frac{2}{1 + \frac{1}{K}}$
$C_{2} = \frac{ε_{0} A}{d} \times \frac{2}{1 + \frac{1}{5}}$
$C_{2} = \frac{5}{3} C$
$∴ %$ increased in capacitance, $= \frac{5}{3} - 1 = \frac{2}{3} \times 100$
$= 66.6 %$
So, increase in the capacitance is $66.6 %$

[JCECE-2009]

Capacitance

166000 The capacitors $A$ and $B$ have identical geometry. A material with a dielectric constant 3 is present between the plates of $B$ . The potential difference across $A$ and $B$ are respectively:

1

2.5 V, 7.5 V

2

2 V, 8 V

3

8 V, 2 V

4

7.5 V, 2.5 V

Explanation:

: Let's potential difference across A and B are $V_{A}$ and $V_{B}$ .
We know that, $Q = CV$
Then, $C \propto \frac{1}{V}$
And capacitance,
$C_{A} = C$
$C_{B} = 3 C$
$\frac{V_{A}}{V_{B}} = \frac{C_{B}}{C_{A}}$
$\frac{V_{A}}{V_{B}} = \frac{3 C}{C}$
$V_{A} = 3 V_{B}$
$(∵ K = 3)$
We know that,
$\begin{array}{ll} V_{A} + V_{B} = 10 \\ 3 V_{B} + V_{B} = 10 \\ 4 V_{B} = 10 \Rightarrow V_{B} = 2.5 V \\ ∴ V_{A} = 3 \times 2.5 = 7.5 V \end{array}$

[JCECE-2005]

Capacitance

166001 A parallel plate capacitor has a capacity $C$ . The separation between the plates is doubled and a dielectric medium is inserted between the plates. If the capacity is $3 C$ , then the dielectric constant of the medium will be

1 1.5

2 3

3 6

4 12

Explanation:

: Given, $C_{1} = C, C_{2} = 3 C$
Let's $d_{1} = d,$ so, $d_{2} = 2 d$
Then, dielectric constant $(K) =$ ?
$C_{1} = \frac{ε_{0} A}{d_{1}} = \frac{ε_{0} A}{d}$
Now, $C_{2} = \frac{K ε_{0} A}{d_{2}}$
$3 C = \frac{K ε_{0} A}{2 d}$
$3 C = \frac{K}{2} \times \frac{ε_{0} A}{d}$
$3 C = \frac{K}{2} C_{1} \Rightarrow 3 C = \frac{K}{2} C$
$K = 6$

[COMEDK 2015]

Question Bank Series

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